Prove that for $f:A \to [0,\infty)$ if $\sum\limits_{a\in\mathbb A} f(a) < \infty$ then $\{a \in A\mid f(a) \neq0\}$ is at most countable.

Question

Prove that for $$f:A \to [0,\infty)$$ if $$\sum\limits_{a\in\mathbb A} f(a) < \infty$$ then $\{a \in A\mid f(a) \neq0\}$ is at most countable.

I tried but i didn't conclude to anything.Any ideas for prove this?

Answer 1

Hint: Consider the sets

$$ A_n := \{a \in A \,\mid \, f(a) \geq 1/n \}. $$

Each of these sets is finite (why?) and $\{a \in A \, \mid \, f(a) \neq 0\} = ???$.

Answer 2

Let $B=\{f(a) : a\in A\}$. Then consider these sets: \begin{align} & B\cap [1,\infty) \\ & B\cap [1/2,1) \\ & B\cap [1/3,1/2) \\ & B\cap [1/4,1/3) \\ & B\cap [1/5,1/4) \\ & \ \ \ \ \ \vdots \end{align} See if you can show that if $\displaystyle\sum_{a\in A} f(a) <\infty $ then every one of the sets above must be finite. And then think about how many sets are in the sequence above.